A spherical oil droplet is released from rest at the bottom of a column of water.
The graph shows the variation with time \(t\) of the vertical velocity of the oil droplet.
(a) The following data are available:
radius of the oil droplet \(=3.5\,\mathrm{mm}\)
weight of the oil droplet \(=1.6\times10^{-3}\,\mathrm{N}\)
density of the water \(=1000\,\mathrm{kg\,m^{-3}}\)
viscosity of the water \(=1.1\times10^{-3}\,\mathrm{Pa\,s}\)
(i) Show that the volume of the oil droplet is about \(2\times10^{-7}\,\mathrm{m^3}\).
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(ii) Calculate the initial acceleration of the oil droplet.
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(b) Describe why the acceleration of the oil droplet changes.
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(c)
(i) Explain why the velocity of the oil droplet is constant for \(t>3\,\mathrm{s}\).
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(ii) Deduce the velocity of the oil droplet for \(t>3\,\mathrm{s}\).
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Most-appropriate topic codes (IB Physics 2025):
(a)(i) Correct Answer: \( \boxed{1.8\times10^{-7}\,\mathrm{m^3}} \)
The volume of a sphere is
\(V=\dfrac{4}{3}\pi r^{3}\)
The radius is \(3.5\,\mathrm{mm}=3.5\times10^{-3}\,\mathrm{m}\).
Therefore,
\(V=\dfrac{4}{3}\pi(3.5\times10^{-3})^{3}\)
\(V\approx1.8\times10^{-7}\,\mathrm{m^3}\)
Hence, \( \boxed{V\approx2\times10^{-7}\,\mathrm{m^3}} \).
(a)(ii) Correct Answer: \( \boxed{0.98\,\mathrm{m\,s^{-2}}} \)
The buoyancy force acting upwards is
\(F_b=\rho Vg\)
\(F_b=(1000)(1.8\times10^{-7})(9.8)\)
\(F_b\approx1.76\times10^{-3}\,\mathrm{N}\)
The mass of the droplet is obtained from its weight:
\(m=\dfrac{W}{g}=\dfrac{1.6\times10^{-3}}{9.8}\)
\(m\approx1.63\times10^{-4}\,\mathrm{kg}\)
At the instant of release, the droplet has zero velocity, so the viscous drag is zero. The net force is therefore
\(F_{\mathrm{net}}=F_b-W\)
\(F_{\mathrm{net}}=1.76\times10^{-3}-1.6\times10^{-3}\,\mathrm{N}\)
Therefore,
\(a=\dfrac{F_{\mathrm{net}}}{m}\)
\(a\approx0.98\,\mathrm{m\,s^{-2}}\)
Thus, the initial acceleration is \( \boxed{0.98\,\mathrm{m\,s^{-2}}} \), in the upward direction.
(b) Correct Answer:
As the speed of the droplet increases, the viscous drag force increases.
The increasing drag force reduces the resultant force, so the acceleration decreases.
(c)(i) Correct Answer:
For \(t>3\,\mathrm{s}\), the droplet has reached terminal velocity.
At terminal velocity, the viscous drag force and weight balance the buoyancy force.
Therefore, the resultant force is zero and hence the acceleration is zero.
With zero acceleration, the velocity remains constant.
(c)(ii) Correct Answer: \( \boxed{2.2\,\mathrm{m\,s^{-1}}} \)
At terminal velocity, Stokes’ law gives the viscous drag force as
\(F_d=6\pi\eta rv_t\)
The forces balance, so
\(F_d=F_b-W\)
Therefore,
\(v_t=\dfrac{F_b-W}{6\pi\eta r}\)
Using \(F_b\approx1.8\times10^{-3}\,\mathrm{N}\), \(W=1.6\times10^{-3}\,\mathrm{N}\), \(\eta=1.1\times10^{-3}\,\mathrm{Pa\,s}\), and \(r=3.5\times10^{-3}\,\mathrm{m}\),
\(v_t=\dfrac{(1.8\times10^{-3})-(1.6\times10^{-3})}{6\pi(1.1\times10^{-3})(3.5\times10^{-3})}\)
\(v_t\approx2.2\,\mathrm{m\,s^{-1}}\)
Hence, the terminal velocity is \( \boxed{2.2\,\mathrm{m\,s^{-1}}} \).
